Problem 40 - Thomas Calculus

Question

Find the function with the given derivative whose graph passes through the point \(P.\) \begin{equation}r^{\prime}(t)=\sec t \tan t-1, \quad P(0,0)\end{equation}

Step-by-Step Solution

Answer

The function is \( r(t) = \sec t - t - 1 \).

Step 1: Understand the Problem

We need to find the original function \( r(t) \) given its derivative \( r'(t) = \sec t \tan t - 1 \), and we know that this function passes through the point \( P(0,0) \).

Step 2: Integrate the Derivative

To find \( r(t) \), we need to integrate the given derivative \( r'(t) \). So we have:\[ r(t) = \int (\sec t \tan t - 1) \, dt \]Split the integral:\[ r(t) = \int \sec t \tan t \, dt - \int 1 \, dt \]

Step 3: Integrate Each Term

Compute the integral of the first term:\[ \int \sec t \tan t \, dt = \sec t + C_1 \] For the second term:\[ \int 1 \, dt = t + C_2 \]Combining these results gives:\[ r(t) = \sec t - t + C \] where \( C = C_1 - C_2 \) is an integration constant.

Step 4: Apply the Initial Condition

Use the point \( P(0,0) \) to find \( C \). Since the function passes through this point:\[ r(0) = \sec(0) - 0 + C = 0 \]\[ \sec(0) = 1 \], therefore:\[ 1 + C = 0 \] leading to \( C = -1 \).

Step 5: Write the Final Solution

Using the value of \( C \), we find the function:\[ r(t) = \sec t - t - 1 \].This is the function whose derivative is \( \sec t \tan t - 1 \) and passes through \( P(0,0) \).

Key Concepts

Derivatives

A derivative gives us the rate of change of a function. It represents how a function's output changes with respect to changes in the input. For example, if we have a function that describes the position of an object over time, the derivative of this function provides us with the velocity of the object.

In this exercise, we start with the derivative of a function, which is given as \( r'(t) = \sec t \tan t - 1 \). Our goal is to reverse this process. By integrating the derivative, we find the original function \( r(t) \). The derivative provides key insight into the behavior of functions, helping us understand critical points where the function may reach local maxima or minima, or points of inflection where the rate of change itself changes.

Calculating derivatives is a fundamental aspect of calculus, as it can be applied to various fields such as physics, engineering, and economics to analyze and predict dynamic systems.

Initial Conditions

Initial conditions are crucial when solving differential equations because they allow us to find a specific solution rather than a general one. In the context of this exercise, the initial condition is given by the point \( P(0,0) \), which signifies that the function \( r(t) \) passes through this point.

When we integrate a derivative, we introduce an arbitrary constant, often denoted as \( C \). This constant represents the family of curves that could be solutions to our differential equation. The initial condition helps us nail down the value of \( C \), which gives us the unique solution that fits our problem.

For the given problem, after integrating the derivative to find \( r(t) = \sec t - t + C \), we apply the initial condition by substituting \( t = 0 \) and \( r(0) = 0 \). This allows us to solve for \( C \) and determines that \( C = -1 \). Thus, initial conditions are essential for obtaining the precise function that behaves exactly as the problem dictates.

Trigonometric Functions

Trigonometric functions are mathematical functions that relate the angles of a triangle to its side lengths. They are primarily used in the context of right-angled triangles, though their utility extends much farther in many areas of mathematics and science.

In this problem, the functions \( \sec t \) and \( \tan t \) appear as part of the derivative. The secant function, \( \sec t \), is the reciprocal of the cosine, so \( \sec t = \frac{1}{\cos t} \). The tangent function, \( \tan t \), is the ratio of the sine to the cosine, i.e., \( \tan t = \frac{\sin t}{\cos t} \).

Trigonometric identities and functions often come into play when solving calculus problems because they describe periodic phenomena, such as waves or circular motions. Understanding how these functions work and their properties can greatly simplify complex calculus problems involving oscillating or circular behavior.

These functions also play a significant role in integrating expressions like \( \sec t \tan t \), as seen in this exercise, where recognizing these derivative forms and identities makes integration much more structured and efficient.